Question

In a parallelogram ABCD, E and F are mid points of side AB ans CD respectively. Show that line segment AF and EC trisect the diagonal.

Answer 1

$\red{\underline{\underline{\sf{Given:}}}}$

Parallelogram ABCD, E and F are midpoints of AB and CD.

[see the image in attached pic]

$\red{\underline{\underline{\sf{To\:\:Prove:}}}}$

AF and EC intersect the diagonal BD

∵ $\boxed{\sf{DG = GH = HB}}$

$\red{\underline{\underline{\sf{Construction:}}}}$

Join AF and EC which meet BD at G and H points

$\red{\underline{\underline{\sf{Proof:}}}}$

AB = CD and AB ║CD        [∴ sides of ║gm]

Now,

$\bf \frac{1}{2}AB = \frac{1}{2}CD \:\:and \:\:\frac{1}{2}AB \parallel \frac{1}{2}CD$

∴ AE = FC and AE║FC

[E and F are midpoints of AB and CD]

∵ AECF is also a parallelogram.

In ΔDHC, F is the midpoint of DC

∴ FG║HC     [opposite line segment of ║gm]

Also, G is the midpoint of DH

∴ DG = GH  ........(i)

Similarly,

In ΔABG

→ E is the midpoint of AB

∴ EH║AG     [Opposite line segment of ║gm]

H is the midpoint of BG

[If a line segment drawn parallel through a midpoint of one side of triangle to the other side of the triangle]

so, GH = HB   .........(ii)

From (i) and (ii)

DG = GH = HB

Hence Proved

∴ AF and EC trisect diagonal BD

Answer 2

Given ABCD is a parallelogram

Hence AB || CD

⇒ AE || FC

Also AB = CD (Opposite sides of parallelogram ABCD)

⇒ AE = FC (Since E and F are midpoints of AB and CD)

In quadrilateral AECF, one pair of opposite sides are equal and parallel.

∴ AECF is a parallelogram.

⇒ AF || EC (Since opposite sides of a parallelogram are parallel)

In ΔDPC, F is the midpoint of DC and FQ || CP

Hence Q is the midpoint of DQ by converse of midpoint theorem.

⇒ DQ = PQ → (1)

Similarly, in ΔAQB, E is the midpoint of AB and EP || AQ

Hence P is the midpoint of DQ by converse of midpoint theorem.

⇒ BP = PQ → (2)

From equations (1) and (2), we get

BP = PQ = DQ

Hence, the line segments AF and EC trisect the diagonal BD of parallelogram ABCD.